Question

Which of the functions is not even?
A) $\log \left( {\dfrac{{1 + {x^2}}}{{1 - {x^2}}}} \right)$
B) ${\sin ^2}x + {\cos ^2}x$
C) $\log \left( {\dfrac{{1 + {x^3}}}{{1 - {x^3}}}} \right)$
D) $\dfrac{{{{(1 + {2^x})}^2}}}{{{2^x}}}$

Answer 1

We will use the odd, even function formula to find which of the options is correct and is not an even function. This is how we will get our answer. Graph of even function is symmetric about y- axis and odd function is symmetric about opposite quadrants.

Complete step by step solution:
Let’s check the option to find the function is not even, let’s check our option A
$\log \left( {\dfrac{{1 + {x^2}}}{{1 - {x^2}}}} \right)$
Now we will apply an odd and even function formula, for that we will compare $f(x)$ by -x in the place of x to see if the $f(x) = f( - x)$ if this condition is satisfied then the function is even. If the condition is not satisfied then the function is odd.
Now lets
$\Rightarrow f(x) = \log \left( {\dfrac{{1 + {x^2}}}{{1 - {x^2}}}} \right)$
Put $x = - x$ in function $f(x)$ which will give us
$\Rightarrow f( - x) = \log \left( {\dfrac{{1 + {{\left( { - x} \right)}^2}}}{{1 - {{\left( { - x} \right)}^2}}}} \right)$
On simplification we get,
$\Rightarrow f( - x) = \log \left( {\dfrac{{1 + {x^2}}}{{1 - {x^2}}}} \right)$
Therefore, the function fulfills the condition
$f( - x) = f(x)$
So, this is an even function
So, option A is not right answer
So, we will move on to the next option to check option B
${\sin ^2}x + {\cos ^2}x$
$\Rightarrow f(x) = {\sin ^2}x + {\cos ^2}x$
On substituting $x = - x$ in function $f(x)$ which will give us
$\Rightarrow f( - x) = {\sin ^2}\left( { - x} \right) + {\cos ^2}\left( { - x} \right)$
Using $\sin ( - x) = - \sin x\,$ and $\cos ( - x) = \cos x\,$ , we get,
$\Rightarrow f( - x) = {( - \sin x)^2} + {(\cos x)^2}$
Since ${(\cos x)^2} = {\cos ^2}x,\,\,{(\sin x)^2} = {\sin ^2}x$ , using this we get,
$\Rightarrow f( - x) = {\sin ^2}x + {\cos ^2}x$
Therefore, the function
$f(x) = f( - x)$
This is an even function
Now we will move on to the option C
$\log \left( {\dfrac{{1 + {x^3}}}{{1 - {x^3}}}} \right)$
$\Rightarrow f(x) = \log \left( {\dfrac{{1 + {x^3}}}{{1 - {x^3}}}} \right)$
On substituting $x = - x$ in function $f(x)$ which will give us

$\Rightarrow f( - x) = \log \left( {\dfrac{{1 + {{\left( { - x} \right)}^3}}}{{1 - {{\left( { - x} \right)}^3}}}} \right)$
On simplification we get,
$\Rightarrow f( - x) = \log \left( {\dfrac{{1 - {x^3}}}{{1 + {x^3}}}} \right)$
Therefore, we get
$\Rightarrow f( - x) \ne f(x)$
By this is we can see the function is not even

Hence the answer is C.

Note:
An even function is symmetric about the y-axis of a graph. An odd function is symmetric about the origin $(0,0)$ of a graph. This means that if you rotate an odd function $180^\circ$ around the origin, you will have the same function you started with. Most functions will be neither even nor odd. The only function that is even and odd is $f\left( x \right) = 0$ . To see if a function is even, you can imagine folding the graph along its y-axis. If the function has folded onto itself, then it is even.